Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant?
Round your answer to three decimal places.

- anonymous

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- anonymous

Okay, you can find the volume first since you have h and r , so :
\[V = \pi r^2 h\]
\[= ( 3)^2 (6) \pi\]
= 169.65 ft^3

- anonymous

sstarica, isn't this the same thing we were tackling

- anonymous

now find V' which is :\[V' = 2\pi rr'h'\]

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## More answers

- anonymous

not really @iam , somehow similar :)

- anonymous

now we have:
h' = 0.2 ft/sec
we need to find r'

- anonymous

do i just plug in

- anonymous

yes :)
but we need to find r' now

- anonymous

\[V' = 2 \pi rr' h'\]
we have h' , but we need to find r' to calculate V' = rate of volume increase

- anonymous

how do i do that? :(

- anonymous

hold on :)

- anonymous

thx

- anonymous

you need to find a relationship between r and h
so r/h = r'/h'

- anonymous

thats where im having problem for all the problems

- anonymous

cross mulitply and you'll get:
r' = rh'/r

- anonymous

sorry! wait you'll get:
r' = rh'/h :)

- anonymous

are you with me soo far?

- anonymous

yes :)

- anonymous

now plug :
r =3 ft
h = 6ft
h' = 0.2ft/sec
r' = (3)(0.2)/6
= 0.1 ft/sec

- anonymous

we have found r' now, let's go back to V'\[V' = 2 \pi rr'h'\]
so,
V' = 2(3)(0.1)(0.2) pi
= 0.377 ft^3/sec

- anonymous

Correct me if I'm wrong please

- anonymous

OMG this makes so much sense now

- anonymous

but the website marked it wrong :(

- anonymous

why?

- anonymous

I guess there's a mistake, but that's the idea, maybe in calculation?

- anonymous

ok

- anonymous

did you get the idea?

- anonymous

one qquestion

- anonymous

sure :)

- anonymous

where did u get \[V \prime = 2\pi r r \prime h \prime \]

- anonymous

you derive with respect to r and h :)

- anonymous

you want to know how is the volume increasing when both the radius and height is changing, that's the story of the related rated problem

- anonymous

oooooh...im sorry can u show me one more problem

- anonymous

hmmmm, let's say we have a circle of radius = 2 m, and the radius is decreasing at 0.5 m/sec. How much will the circle's area increase ?

- anonymous

I'll give you advise, whenever you see ( m/sec) or anything with respect to time it's something' which in this case is the r' = 0.5m/sec

- anonymous

so what is the given?

- anonymous

we have :
- R = 2 m
- R' = 0.5 m/sec
and we know the formula of the Area of the circle which is:
A = pi r^2, right?

- anonymous

uhhuh

- anonymous

Now , you want to know how much is the area increasing with respect to R, since R is increasing :)
so you find the derivative of it which will be :
A' = 2 pi rr'

- anonymous

so when exactly do i start pluging in numbers?

- anonymous

right now, after you find the derived formula, start pluggingthe numbers :)

- anonymous

okay ...these are pretty simple compared to the ones i have on hw :'(

- anonymous

Like this problem, they want to know how much is the area increasing , we got the formula so the answer is:
A' = 2(2)(0.5) pi
= 2 pi = 2.68 m^2/sec :)
So the area is increasing at 2.68 m^2/sec

- anonymous

Lol, when you understand the small idea, you'll be able to figure out the bigger idea :)

- anonymous

did you understand my problem?

- anonymous

yes but how would i approach towards a hemisphere problem

- anonymous

think about it, write down the given on one side, and the RTF(required to find) on the other side, then SKETCH!
To test your understanding , you've got to sketch what they give you to be fully understand what they want :)

- anonymous

to fully*

- anonymous

1) write the given and what you need to find
2) sketch
3) write down the general formula that you're going to use :)

- anonymous

^_^

- anonymous

ok :)

- anonymous

good luck !
You're a smart student, you can figure it out :)

- anonymous

you got the small idea, you'll be able to solve any bigger ideas now . They are not even big, they are small ideas that got stretched ^_^, don't let the problem boss u, YOU ARE THE BOSS!

- anonymous

lol..thx

- anonymous

np :)
Give it a try once more. ^_^

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